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Sagot :
Certainly! Let's proceed with the detailed, step-by-step solution for finding the center and the radius of the circle given by the equation [tex]\( x^2 + 4x + y^2 - 6y = -4 \)[/tex].
Step 1: Understand the general form of a circle’s equation
The standard form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Step 2: Rewrite the given equation by completing the square
We start with the equation:
[tex]\[ x^2 + 4x + y^2 - 6y = -4 \][/tex]
Completing the square for the [tex]\(x\)[/tex]-terms:
Take the quadratic term in [tex]\(x\)[/tex], which is [tex]\( x^2 + 4x \)[/tex].
To complete the square, follow these steps:
1. Take the coefficient of [tex]\(x\)[/tex], which is [tex]\(4\)[/tex].
2. Divide it by [tex]\(2\)[/tex], giving [tex]\(\frac{4}{2} = 2\)[/tex].
3. Square the result, giving [tex]\(2^2 = 4\)[/tex].
So, [tex]\(x^2 + 4x\)[/tex] can be rewritten as:
[tex]\[ x^2 + 4x = (x + 2)^2 - 4 \][/tex]
Completing the square for the [tex]\(y\)[/tex]-terms:
Take the quadratic term in [tex]\(y\)[/tex], which is [tex]\( y^2 - 6y \)[/tex].
To complete the square, follow these steps:
1. Take the coefficient of [tex]\(y\)[/tex], which is [tex]\(-6\)[/tex].
2. Divide it by [tex]\(2\)[/tex], giving [tex]\(\frac{-6}{2} = -3\)[/tex].
3. Square the result, giving [tex]\(-3^2 = 9\)[/tex].
So, [tex]\(y^2 - 6y\)[/tex] can be rewritten as:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]
Step 3: Substitute the completed squares back into the original equation
Substitute these completed squares into the original equation:
[tex]\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 = -4 \][/tex]
Combine the constants on the right-hand side:
[tex]\[ (x + 2)^2 + (y - 3)^2 - 13 = -4 \][/tex]
[tex]\[ (x + 2)^2 + (y - 3)^2 = 9 \][/tex]
Step 4: Identify the center and the radius
Now, the equation is in the standard form of a circle:
[tex]\[ (x - (-2))^2 + (y - 3)^2 = 3^2 \][/tex]
From this, we can determine:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\((-2, 3)\)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\( \sqrt{9} = 3 \)[/tex]
Final Answer:
The center is located at [tex]\((-2, 3)\)[/tex], and the radius is [tex]\(3\)[/tex].
Step 1: Understand the general form of a circle’s equation
The standard form of a circle's equation is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]
where [tex]\((h, k)\)[/tex] is the center of the circle and [tex]\(r\)[/tex] is the radius.
Step 2: Rewrite the given equation by completing the square
We start with the equation:
[tex]\[ x^2 + 4x + y^2 - 6y = -4 \][/tex]
Completing the square for the [tex]\(x\)[/tex]-terms:
Take the quadratic term in [tex]\(x\)[/tex], which is [tex]\( x^2 + 4x \)[/tex].
To complete the square, follow these steps:
1. Take the coefficient of [tex]\(x\)[/tex], which is [tex]\(4\)[/tex].
2. Divide it by [tex]\(2\)[/tex], giving [tex]\(\frac{4}{2} = 2\)[/tex].
3. Square the result, giving [tex]\(2^2 = 4\)[/tex].
So, [tex]\(x^2 + 4x\)[/tex] can be rewritten as:
[tex]\[ x^2 + 4x = (x + 2)^2 - 4 \][/tex]
Completing the square for the [tex]\(y\)[/tex]-terms:
Take the quadratic term in [tex]\(y\)[/tex], which is [tex]\( y^2 - 6y \)[/tex].
To complete the square, follow these steps:
1. Take the coefficient of [tex]\(y\)[/tex], which is [tex]\(-6\)[/tex].
2. Divide it by [tex]\(2\)[/tex], giving [tex]\(\frac{-6}{2} = -3\)[/tex].
3. Square the result, giving [tex]\(-3^2 = 9\)[/tex].
So, [tex]\(y^2 - 6y\)[/tex] can be rewritten as:
[tex]\[ y^2 - 6y = (y - 3)^2 - 9 \][/tex]
Step 3: Substitute the completed squares back into the original equation
Substitute these completed squares into the original equation:
[tex]\[ (x + 2)^2 - 4 + (y - 3)^2 - 9 = -4 \][/tex]
Combine the constants on the right-hand side:
[tex]\[ (x + 2)^2 + (y - 3)^2 - 13 = -4 \][/tex]
[tex]\[ (x + 2)^2 + (y - 3)^2 = 9 \][/tex]
Step 4: Identify the center and the radius
Now, the equation is in the standard form of a circle:
[tex]\[ (x - (-2))^2 + (y - 3)^2 = 3^2 \][/tex]
From this, we can determine:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\((-2, 3)\)[/tex]
- The radius [tex]\(r\)[/tex] is [tex]\( \sqrt{9} = 3 \)[/tex]
Final Answer:
The center is located at [tex]\((-2, 3)\)[/tex], and the radius is [tex]\(3\)[/tex].
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